Integrate
e^(2x)cosx dx
I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?
Integrate
e^(2x)cosx dx
I tried substitution, where u=2x, but then I am still left with an x value from cosx. I know how to integrate them separately, but now when they are multipled together... how should I go about integrating?
You need to use integration by parts.
$\displaystyle \int{u\,dv} = u\,v - \int{v\,du}$.
So for $\displaystyle \int{e^{2x}\cos{x}\,dx}$
Let $\displaystyle u = e^{2x}$ so that $\displaystyle du = 2e^{2x}$.
Let $\displaystyle dv = \cos{x}$ so that $\displaystyle v = \sin{x}$.
Therefore $\displaystyle \int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} - \int{2e^{2x}\sin{x}\,dx}$
$\displaystyle = e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx}$.
Now you need integration by parts again.
Let $\displaystyle u = e^{2x}$ so that $\displaystyle du = 2e^{2x}$.
Let $\displaystyle dv = \sin{x}$ so that $\displaystyle v = -\cos{x}$.
So $\displaystyle e^{2x}\sin{x} - 2\int{e^{2x}\sin{x}\,dx} = e^{2x}\sin{x} - 2\left[-e^{2x}\cos{x} - \int{-2e^{2x}\cos{x}\,dx}\right]$
$\displaystyle = e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}$.
So we have
$\displaystyle \int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x} - 4\int{e^{2x}\cos{x}\,dx}$
$\displaystyle 5\int{e^{2x}\cos{x}\,dx} = e^{2x}\sin{x} + 2e^{2x}\cos{x}$
$\displaystyle \int{e^{2x}\cos{x}\,dx} = \frac{1}{5}e^{2x}\sin{x} + \frac{2}{5}e^{2x}\cos{x} + C$.